Rotational Dynamics Physics 1 Sir Lost-a-Lot dons his armor and sets out from th

Question

Rotational Dynamics Physics 1

Sir Lost-a-Lot dons his armor and sets out from the castle on his trusty steed in his quest to improve communication between damsels and dragons. Unfortunately his squire lowered the draw bridge too far and finally stopped it 20.0° below the horizontal. Lost-a-Lot and his horse stop when their combined center of mass is 1.00 m from the end of the bridge. The uniform bridge is 7.50 m long and has a mass of 2400 kg. The lift cable is attached to the bridge 5.00 m from the hinge at the castle end, and to a point on the castle wall 12.0 m above the bridge. Lost-a-Lot's mass combined with his armor and steed is 950 kg. Hint: First determine all the angles and lengths of the triangle made by the wall, the cable, and the drawbridge.

a) Determine tension of cable___

b) Magnitude of the horizontal force
Direction_____

to the left or to the right?

c) Magnitude of the vertical force
Direction _____

down or up ?

Explanation / Answer

We need the angle the cable makes with the bridge.

The angle between the bridge and the wall (vertical) is 70º.
The law of cosines gives us
C² = 5² + 12² - 2*5*12*cos70º = 128
C = 11.3 m length of cable

Now we can use the law of sines to get the angle between the bridge and cable:
sin / 12m = sin70º / 11.3m
sin = 0.997
= arcsin0.997 = 85.46º

(a) Now sum the moments about the hinge:
M = 0 = (2400kg *(½)*7.5m + 950kg * 6.5m)*9.8m/s²*cos20º - T*sin(85.46º)*5m
where T is the cable tension. Solving, find
T = 139746N·m / (5m*sin85.46) = 139746N-m /4.98m = 28061 N

(b) ß = 180º - 70º - 85.46º = 24.54º angle between cable and wall
horizontal force Fx = T*sin(24.54) = 28061N*0.4153 11654 N
away from wall

(c) vertical force Fy = (2400 + 950)kg * 9.8m/s² - T * cos24.54

                                  = (2400 + 950)kg * 9.8m/s² - 28061N * cos24.54 = 7303.7 N
Magnitude of the vertical force direction up.

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